Quantitative Aptitude Formulas for Placement Tests

The quantitative sections of placement aptitude tests draw on a familiar set of topics, and many of their questions reduce to a few formulas. This sheet collects those formulas by topic, with one worked example each so you can see them applied.

Formulas save time only when you know why they work. Try to derive each one once from first principles: if you forget it under time pressure, you can still reason your way to the answer. Throughout, × means multiply, / means divide and ^ means "to the power of", so (1 + R/100)^n is (1 + R/100) multiplied by itself n times.

Percentages

The same idea covers price and consumption: if a price rises by x%, consumption must fall by x/(100 + x) × 100 % to keep spending unchanged. If the price falls by x%, consumption can rise by x/(100 − x) × 100 %.

Knowing common fractions as percentages speeds up most of these calculations (values rounded to two decimal places):

Worked example. The price of sugar rises by 25%. By what percentage must a family reduce its consumption so that its spending stays the same?

Reduction = 25/(100 + 25) × 100 = 25/125 × 100 = 20%. Check: at a price of 100 per kg, 1 kg costs 100. At the new price of 125 per kg, 0.8 kg costs 100, which is 20% less sugar.

Profit, loss and discount

CP is the cost price, SP the selling price and MP the marked (list) price. Profit and loss percentages are calculated on CP; discount percentages are calculated on MP.

Worked example. A shopkeeper marks goods 40% above cost price and offers a 10% discount. What is the profit percentage?

Using m − d − m × d/100: 40 − 10 − (40 × 10)/100 = 40 − 10 − 4 = 26%. Check with CP = 100: MP = 140, SP = 140 × 0.9 = 126, so the profit is 26 on a cost of 100.

Simple and compound interest

P is the principal, R the rate per cent per year, T or n the time in years, and A the amount (principal plus interest).

At compound interest, if a sum becomes x times itself in n years, it becomes x² times in 2n years and x³ times in 3n years.

Worked example. Find the difference between compound and simple interest on ₹5,000 for 2 years at 8% a year.

Using P × (R/100)²: 5,000 × (8/100)² = 5,000 × 0.0064 = ₹32. Check: SI = 5,000 × 8 × 2/100 = ₹800. The compound amount is 5,000 × 1.08² = 5,000 × 1.1664 = ₹5,832, so CI = ₹832, and 832 − 800 = ₹32.

If a question does not say how interest is compounded, assume yearly. Convert months to years before substituting (9 months is 0.75 years), and for half-yearly or quarterly compounding change both the rate and the number of periods.

Ratio, proportion and averages

Ratio and proportion

Averages and mixtures

Worked example. The average age of 30 students is 15 years. When the teacher's age is included, the average rises by 1 year. How old is the teacher?

New member = 15 + (30 + 1) × 1 = 46 years. Check: the students' ages total 30 × 15 = 450. With the teacher, the total is 31 × 16 = 496, and 496 − 450 = 46.

Time and work

If someone completes a job in a days, they do 1/a of it each day. Add rates, not days.

In the team formula, M is the number of people, D the days, H the hours per day and W the amount of work. When people are paid jointly for a job, the payment is split in the ratio of the work each one did.

A faster method than fractions: take the total work as the LCM of the individual times. Each person's daily work is then a whole number of units.

Worked example. A can finish a job in 12 days and B in 18 days. They work together for 4 days, then A leaves. How many more days does B need?

Take the total work as LCM(12, 18) = 36 units. A does 36/12 = 3 units a day and B does 36/18 = 2. In 4 days together they complete 4 × 5 = 20 units, leaving 16. B needs 16/2 = 8 more days.

Pipes and cisterns

These are time and work problems where emptying counts as negative work. An inlet that fills a tank in a hours adds 1/a of the tank per hour; an outlet that empties it in b hours removes 1/b.

Worked example. Pipes A and B fill a tank in 20 and 30 minutes, and pipe C empties it in 15 minutes. If all three are opened on an empty tank, how long does it take to fill?

Take the tank as LCM(20, 30, 15) = 60 units. A adds 3 units a minute, B adds 2 and C removes 4, so the net rate is 3 + 2 − 4 = 1 unit a minute. The tank fills in 60 minutes.

Speed, distance and time

Here x and y are speeds, L lengths and t times.

Worked example. A 200 m train running at 72 km/h crosses a 250 m platform. How long does it take?

72 km/h = 72 × 5/18 = 20 m/s. The train must cover its own length plus the platform: 200 + 250 = 450 m. Time = 450/20 = 22.5 seconds.

Boats and streams

u is the boat's speed in still water and v the speed of the stream, with u > v.

Worked example. A boat covers 24 km downstream in 2 hours and returns in 3 hours. Find its speed in still water and the speed of the stream.

Downstream speed = 24/2 = 12 km/h and upstream speed = 24/3 = 8 km/h. Speed in still water = (12 + 8)/2 = 10 km/h, and the stream's speed = (12 − 8)/2 = 2 km/h.

Permutations and combinations

Use permutations when order matters (arrangements, rankings, codes) and combinations when it doesn't (teams, committees, selections).

When a task happens in steps (this and then that), multiply the counts. When there are separate cases (this or that), add them.

Worked example. A committee of 3 men and 2 women is chosen from 6 men and 5 women. How many different committees are possible?

Men: 6C3 = 6!/(3! × 3!) = 20. Women: 5C2 = 5!/(2! × 3!) = 10. Both choices must happen, so multiply: 20 × 10 = 200.

Probability

For equally likely outcomes, P(E) = favourable outcomes/total outcomes, which always lies between 0 and 1.

Mutually exclusive and independent are different ideas: if two events each have a non-zero chance, they cannot be both.

Sample spaces worth remembering:

• Coins. n coins give 2^n outcomes.
• Dice. One die gives 6 outcomes, two dice 36, three dice 216. With two dice, a sum of s can be made in s − 1 ways for s from 2 to 7, and 13 − s ways for s from 7 to 12.
• Cards. A standard deck has 52 cards in 4 suits of 13. There are 26 red cards (hearts and diamonds), 26 black (spades and clubs), 12 face cards (jack, queen and king of each suit) and 4 aces.

Worked example. Two cards are drawn at random, without replacement, from a standard deck. What is the probability that both are aces?

Ways to choose 2 aces: 4C2 = 6. Ways to choose any 2 cards: 52C2 = 52 × 51/2 = 1,326. Probability = 6/1,326 = 1/221. Check the other way: (4/52) × (3/51) = 12/2,652 = 1/221.

Number system

Divisibility rules

For other divisors, split them into co-prime factors and test each: a number is divisible by 15 if it is divisible by 3 and by 5. For example, 364 is divisible by 7 because 36 − 2 × 4 = 28.

HCF and LCM

For example, the smallest number above 3 that leaves remainder 3 when divided by 6, 8 and 12 is LCM(6, 8, 12) + 3 = 24 + 3 = 27.

Remainders and last digits

• dividend = divisor × quotient + remainder, where the remainder is less than the divisor.
• The remainder of a sum or product equals the remainder of the sum or product of the individual remainders. For 23 × 17 divided by 5, the remainders are 3 and 2, and 3 × 2 = 6 leaves 1, so 391 leaves remainder 1.
• For a whole number a greater than 1, (a + 1)^n leaves remainder 1 when divided by a.
• For the same a, (a − 1)^n divided by a leaves remainder 1 when n is even, and a − 1 when n is odd.
• a^n − b^n is divisible by a − b for every n, and by a + b when n is even. a^n + b^n is divisible by a + b when n is odd.
• Fermat's little theorem: if p is prime and a is not a multiple of p, a^(p − 1) leaves remainder 1 when divided by p.

The last digit of a power depends only on the last digit of the base, and repeats in a cycle:

For a cycle of 4, divide the exponent by 4 and use the remainder as the position in the cycle; a remainder of 0 means the fourth position.

Factors, trailing zeros and series

• If N = p^a × q^b × r^c with p, q and r prime, N has (a + 1) × (b + 1) × (c + 1) factors.
• The sum of those factors is (p^(a+1) − 1)/(p − 1) × (q^(b+1) − 1)/(q − 1) × (r^(c+1) − 1)/(r − 1).
• The number of zeros at the end of n! is the sum of the whole-number parts of n/5, n/25, n/125 and so on. For 100!, that is 20 + 4 = 24.
• Sum of the first n natural numbers: n × (n + 1)/2. Of their squares: n × (n + 1) × (2n + 1)/6. Of their cubes: (n × (n + 1)/2)².
• Sum of the first n odd numbers: n². Of the first n even numbers: n × (n + 1).

Worked example. Find the remainder when 3^21 is divided by 5.

Powers of 3 end in 3, 9, 7, 1, in a cycle of 4. 21 divided by 4 leaves remainder 1, so 3^21 ends in 3, and any number ending in 3 leaves remainder 3 when divided by 5. Check with Fermat's little theorem: 3^4 leaves remainder 1 when divided by 5, so 3^20 = (3^4)^5 also leaves 1, and 3^21 = 3^20 × 3 leaves 1 × 3 = 3.

Squares up to 30, cubes up to 15 and the fraction table in the percentages section are worth knowing by heart. They turn many calculations into recognition, which saves time in every topic on this sheet.

Before the test

A few checks catch a lot of avoidable mistakes:

• Units. Convert between km/h and m/s, months and years, and minutes and hours before substituting.
• The base of a percentage. Profit and loss are on cost price, discount is on marked price, and "A is x% more than B" uses B as the base.
• Average speed. Total distance over total time, never the average of the speeds.
• Order. Decide whether order matters before choosing between permutations and combinations.
• "At least one". Usually fastest as 1 − P(none).
• Options. Estimate the answer before calculating, and rule out options that are clearly too large or too small.
• The final quantity. Many wrong answers are correct calculations of a different quantity, such as compound interest instead of the amount, or the stream's speed instead of the boat's.